Ta có : \(x^2\) - 3x = 0
\(\Rightarrow\) x. ( x - 3 ) = 0
\(\Rightarrow\)\(\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
Vậy x = 0; x=3
a) -5 (2x + 6 ) +b | 3x + 9 | = 7x
b) 7(x - 4 ) - | 4x - 12 | = 0
c) | x - 5 | - | x - 3 | =0
d)| x + 1| + | -x + 12 | = 2x -7
e) | x + 2| - | x + 6 | - (-15) = 0
f) | 2x- 10 | + | 3y - 4 | = 0
g) | - 3x + 1 | - | 2x + 8 | + 3x = 0
h) 3( x - 1 ) - 5 | x + 3 | = 0
1/3x+2/5(x-1)=0
a, X+5=2015-(12-7) c, (105-x):25=20150+1
b,-7|x+3|=-49 d,10-2x=25-3x
1. Find the positive value of x such that:
\(x^2-2-2x-2\left|x-1\right|=0\)
2. Find the remainder of the division:
\(\left(x^3-13+5x-3x^2\right):\left(x-3\right)\)
Tìm x , y \(\in\) Z biết
a, xy + 2x + 2y = -16
b, ( x - 3 ) . ( y + 2 ) = 11
c, xy - 3x - y = 0
\(x^3\left(x\text{-}5\right)+3x\left(x\text{-}5\right)=0\)
Tìm x \(\in\) Z biết ( x + 5 ) . ( 3x - 12 ) > 0
Tìm x biết:
a. -2/3.x+1/5=3/10/3x-1/2=1/10
b. |x|=3/4=5/3
c. 2/3x-1/2=1/10
d. 3/5+4/9:x=2/3
e. |x+75phần trăm|=11/5
f. (x+1/2).(2/3-2x)=0
g. 2/3x-3/2x=5/12
h. |2x-1/3|+5/6=1
i. x+2/5=7/12-5/4
j. (14/5x-50):2/3=51
k. 2/5+3/5.(3x-3,7)=-53/10
l. 7/9:(2+3/4x)+5/9=23/27
Giải phương trình nghiệm nguyên:
\(x^2+2y^2-2xy+4x-3y-26=0\)
\(\sqrt{9x^2+16x+96}=3x-16y-24\)
E cần gấp ạ!
